Suppose that $f : X \rightarrow Y$ is mapping between topological spaces that is not continuous at $x_0$. Then there is an open set $V$ in $Y$ containing $f(x_0)$ such that for any open set $U$ containing $x_0$, there is some $x_U \in U$ with $f(x_U) \notin V$. By picking one from each $U$ we can build a net $x_U$ converging to $x_0$ such that $f(x_U)$ does not converge to $f(x_0)$. This however requires the axiom of choice because we have to pick $x_U$ from each $U$.

My question is: if we are given that a function between topological spaces is continuous if and only if it is "sequentially" continuous (in the sense of nets, not necessarily infinite sequences indexed by the integers) - does the axiom of choice follow?

1 Answer
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The axiom of choice is not needed if $X$ is $T_1$; i.e., singleton sets are closed.

Suppose $f:X\to Y$ is not continuous at $x_0$. Let $D$ be the collection of all $(U,x)$ s.t. $U$ is an open set containing both $x_0$ and $x$ and $x_0\not=x$. Say that $(U,x) \le (V,y)$ provided $V\subset U$ and $x$ is not in $V$. If $X$ is $T_1$, this is a directed set. Define a net $g$ with domaine $D$ by $g(U,x)=x$. Then $g$ converges to $x_0$ but $f(g)$ does not converge to $f(x_0)$.

This is a nice example where you avoid picking an element (maybe because you just cannot, in the absence of choice!) and solve the problem by using all possible choices---like someone wanted in another question.
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Mariano Suárez-Alvarez♦Dec 25 '10 at 21:53

This is a good answer. I am still interested in the other direction. The proof that Tychonoff's theorem, or that every vector space has a basis imply the axiom of choice - I find very fascinating and I wonder if similar techniques can be applied in this case.
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mathahadaDec 26 '10 at 14:35

@mathahad: The usual proof that $f$ is continuous at $x_0$ implies that $f(g)$ converges to $f(x_0)$ whenever $g$ is a net converging to $x_0$ does not use the axiom of choice. Maybe you mean something different?
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Bill JohnsonDec 26 '10 at 18:29

I know. My question is if the following are equivalent: 1. Axiom of choice 2. For every mapping $f : X -> Y$ that is not continuous at $x_0$ there is a net converging to $x_0$ that does not converge to $f(x_0)$ It is not hard to prove that 1 implies 2. As you have shown, to prove 2 you don't even have to assume 1 if the space is $T_1$ (the question however is not settled for arbitrary topological spaces)
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mathahadaDec 28 '10 at 8:48

Actually you do not need $T_1$. For the directed set, take the same underlying set, but order it differently: $(U,x)< (V,y)$ provided $U$ is a proper subset of $V$. The net $g$ is the same. This order gives a directed set as long as $x_0$ does not have a smallest open neighborhood, which is a case that is easy to treat separately.
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Bill JohnsonDec 28 '10 at 18:58